SOLUTIONS TO PS4
Xiaoguang Ma
Solution/Proof of Problem 1.
Consider the open set
�
B n
= ( x
1
, x
2
) ∈
R
2
: x
2
1
+ x
2
2
1 �
< 1 − .
n
Then can we find can an n see big that E enough
⊂ ∪ B such n because for any point ( x, y ) ∈ E , x 2 that x 2 + y 2 < 1 − 1 n
, i.e.
( x, y ) ∈ B n
.
+ y 2
It is easy to see there is no finite subcover.
< 1 , we
Solution/Proof of Problem 2.
At first, from the definition, we have d ( x, x ) = 0 .
From the equality d ( x, y ) = || x || + || y || = || y || + || x || = d ( y, x ) , we have d ( x, y ) = d ( y, x ) .
We also have n
� d ( x, y ) = || x || + || y || = ( x
2 i
)
1 / 2 n
�
+ ( y
2 i
)
1 / 2 ≥ 0 i =1 i =1 and it is easy to see that d ( x, y ) = 0 iff x = y = 0 .
From
|| x || + || y || ≤ || x || + || z || + || z || + || y || we have d ( x, y ) ≤ d ( x, z ) + d ( y, z ) .
So d is a metric on
R n .
Open set in (
R k open ball in (
R k
, d
, d
),
) may be not open in (
R k
, d
Euclid
).
For example, consider the
B r
( x ) = { y ∈
R k
: d ( x, y ) < r } .
When r > || x || , then we have
B r
( x ) = { y ∈
R k
: d ( x, y ) < r } = { y ∈
R k
: || x ||| + || y || < r }
= { y ∈
R k
: || y || < r − || x ||} , is just the open ball B r −|| x ||
(0) under the Euclidean metric.
But when r < || x || , then we have
B r
( x ) = { y ∈
R k
: d ( x, y ) < r } = { y ∈
R k
: | x ||| + || y || < r } = { x } , is not open under the Euclidean metric.
Conversely, consider an open set U ⊂ (
R k
, d
Euclid
).
Since for any point x = 0, under the new metric, B r
( x ) = { x } ⊂ U for any r < || x || ; if x = 0, then B r
(0) ⊂
R k for some small r under the new metric.
So we can always find an open neighborhood of x in the new metric that is contained in U .
This means U is open in the new metric.
1

2 SOLUTIONS TO PS4
Solution/Proof of Problem 3.
First, recall that in any metric space a finite set is compact.
We will show that for the discrete metric, these are the only compact sets.
Notice that every subset { x } which contains only one point in X is an open subset.
Indeed, if Y = { x } , then, for any r < 1 , we have B r
( x ) = { x } hence
B r
( x ) ⊂ Y , i.e., Y is open.
Now suppose Y is a compact subset of X .
We can consider an open cover
Y ⊂ y ∈ Y
U y
, where U
So Y ⊂ finitely many y ∈ Y
U y y
.
= { y } .
Since Y is compact, this cover has a finite subcover.
Hence Y is a finite set.
Solution/Proof of Problem 4.
From the definition of E , we can see that E ⊂
{ x ∈
Q
, − 3 < x < 3 } .
So it is bounded.
E is closed .
Recall that, in any metric space, a set E is closed if and only if its complement is open.
If x is any point whose square is less than 2 or greater than
3 then it is clear that there is a nieghborhood around x that does not intersect E .
Indeed, take any such neighborhood in the real numbers and then intersect with the rational numbers.
So the only problem would be at points whose square is exactly 2 or 3 , but we know that are no such points within the rational numbers.
E is not compact .
Consider the open cover
U n
= { x ∈
Q
: 2 + 1 /n < x
2
< 3 − 1 /n } , n ∈
N
.
It is easy to see that it has no finite subcover.
E is open .
Given any point x ∈ E there is a neighborhood of x within the real numbers of elements whose square is between 2 and 3 , intersect this with the rational numbers to see that E is an open subset of
Q
.
Solution/Proof of Problem 5.
Suppose X and Y are two compact sets.
If { U
α
} is an open cover for X ∪ Y , then it is also an open cover of X .
Since X is compact, there is a finite subcover { U
β
}
β ∈ I also have a finite subcover { U
β
}
β ∈ J
⊂ { U
α
} which still covers X .
Similarly, we
⊂ { U
α
} which covers Y .
Putting these covers together,
{ U
β
}
β ∈ I ∪ J
⊂ { U
α
} , we get a finite subcover of X ∪ Y .
So by the definition, X ∪ Y is compact.
Since X is a compact set in a metric space, it is closed.
Hence X ∩ Y is the intersection of a closed set with a compact set.
From Theorem 2.35’s corollary, we can see that X ∩ Y is compact.
Solution/Proof of Problem 6.
The set { 1 } has no limit points because any neighborhood of this point has only one element 1 .
The statement can be proved as the follows.
Let { x k
} ∞ k =1 be a convergent sequence in a metric space with infinitely many distinct elements.
Suppose lim x n n →∞
= x
0
.
Then by the definition of the limits, for any neighborhood of x
0
, there are at most finitely many points in the sequence outside the neighborhood.
So we can always choose a point different from x
0 point of the set.
in any neighborhood which means x
0 is a limit

SOLUTIONS TO PS4 3
Solution/Proof of Problem 7.
Because they are countable, it is possible to put the rational numbers in [0 , 1] in a sequence, ( p n
) .
We claim that every point x ∈
[0 , 1] is a limit of a subsequence of ( p n
) .
We proved in class that between any two real numbers there is a rational number, it follows that between any two real numbers there are infinitely many rational numbers.
This allows us to construct a subsequence of p n
Assume for simplicity that x and x +
1
10 are both in [0 , 1] .
converging
From to among x as the follows.
infinitely many fall in rational x < p n
1 numbers
≤ x + between x and x +
1
10
, let p n
1 be the first of the p n to
1
10
.
Because there are inifinitely many rational numbers between x and p p n n
1 ocurring after p we can pick p n
2 n
1 to be the first rational number in the sequence to fall inside x < p n
2
≤ x +
1
10 2
.
Similarly, we choose p n
3 to be the first rational number in the sequence p n x < p n
3
≤ x +
1
10 3
.
which we denote p n k
Continuing with the in this property fashion, that ocurring we after achieve a p n
2 to fall subsequence inside of p n
, x < p n k
≤ x +
1
10 k
, for any k ∈
N
.
Hence p n
→ x .
If
This that is k was not done the under case, the but assumption x < 1 then that we can both find x
N and x such
+
1
10 that were x and both x + in
1
[0
10 N
, 1] .
are both in [0 , 1] and we can start the construction from there.
Finally, if x = 1 then x and that x p
− n k
1
10 are satisfy both x − in
1
10 k
[0 ,
≤
1] p and n k we
< x can for carry every k .
out the
Hence in every case we obtain a subsequence of ( p n above
) that construction converges to requiring x .
Solution/Proof of Problem 8.
The question is asking: If A is connected, does the interior of A have to be connected?
does the closure of A have to be connected?
Closure of a connected set is always connected.
Suppose E = A ∪ B , where
A ∩ B = ∅ and A ∩ B = ∅ , we show that E is connected by proving that either A or
B must be empty.
We know that E is connected and E = ( A ∩ E ) ∪ ( B ∩ E ) with A ∩ E , B ∩ E separated sets, hence we must have A ∩ E = ∅ or B ∩ E = ∅ .
Say that A ∩ E = ∅ , then E ⊆ B and hence E ⊆ B .
But we know that A ∩ B = ∅ , hence
A = A ∩ ( A ∪ B ) = A ∩ E ⊆ A ∩ B = ∅ , which implies that E is connected.
The interior of a connected set may not be connected.
Consider two tangent closed disk in
R
2 : B
1
((0 , 1)) and B
1
((0 , − 1)) .
The union will give us a connected set.
But the interior part of it will be two separated open balls.

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